Past exam of the mathematics course of the University of Cambridge / 2021 / ii / Paper 3 / 22H (Analysis of Functions)

The Riemann-Lebesgue lemma states that the Fourier transform of an $L^1$ function is continuous and tends to zero at infinity.

For a Schwartz function $f$, differentiation under the integral and integration by parts giveThe Riemann-Lebesgue lemma bounds the right side by an $L^1$ seminorm of $D^{\alpha }(x^{\beta }f)$. Every Schwartz seminorm of $\hat{f}$ is therefore bounded by finitely many Schwartz seminorms of $f$. Hence the Fourier transform maps $\mathcal{S}({\mathbb{R}}^n)$ continuously into itself.

Taking the Fourier transform of a derivative givesThus, away from $\xi =0$,This formula determines the $L^2$ solution uniquely; a distribution supported only at $\xi =0$ cannot belong to $L^2$ unless it is zero.

For $|\xi |\le 1$, the Fourier transform estimate $|\hat{f}(\xi )|\le C\Vert f{\Vert }_1$ and local integrability of $|\xi {|}^{-2}$ in three dimensions control the squared norm. For $|\xi |>1$,The Plancherel theorem therefore yields $\Vert B_j{\Vert }_{H^{s+1}}\le C(\Vert f{\Vert }_1+\Vert f{\Vert }_{H^s})$. Finally the Sobolev embedding theorem in three dimensions gives $H^{s+1}\subset C^1$ when $s+1>1+3/2$, namely when $s>3/2$.